Escaping from a circle of fat lions. You are surrounded, by X fat lions equally spaced around a circle of radius 200 meters in an open field.  While making your escape plan you note several things: they are slow, they can only travel at one tenth of your speed, they are stupid, 
they can only move directly at your current position, and they can’t cooperate with one another. If any lion gets within 1 meter of you, you will be eaten.
What is the maximum value of X for which you have a strategy to escape from them?
 A: TL;DR  Assuming @SimpleArt's strategy of running straight towards the midpoint in between two adjacent lions, one can escape from $364$ lions, but will be caught by $365$. The result is derived from the actual pursuit curve, with the final equation having to be solved numerically. (This is close to, albeit slightly lower than, the value of $366$ proposed by @Jens in a comment.)
Assume oneself (= target) at the origin of the euclidian plane, darting upwards along the positive $y$ axis towards the midpoint between two lions symmetric to the axis. The phases of the ensuing pursuit are described in 
the following (based on the animation at Wolfram's pursuit curve page).



*

*(1) Pursuit starts at $t=0$ with the lion staring down at the origin.

*(2) At each moment in time the lion aims directly towards its target.

*(3) Since the pursuer goes downwards while the target moves upwards, they will be level at some point, after which the pursuer will turn upwards as well. If at this $t=t_1$ the abscissa of the lion is less than its reach $x_1 \le 1$ then the gap is closed and the target has been captured.

*(4) Otherwise the target has escaped, and the lion will continue chasing it pointlessly.
The above suggests the following strategy to solving the problem:


*

*(a) derive the equation of the pursuit curve as a function $y=f(x)$, then

*(b) determine the turning point $f'(x_1)=0$, and finally

*(c) compare its abscissa $x_1$ to the lion reach of $1$ meter; in particular, solving the limit case $x_1=1$ gives the condition for the maximum number of lions that still allows escaping.


The math is quite similar to Pursuit curves solution (with the necessary adjustments), so the following will be skimpier on step-by-step explanations.
Let the speeds of the target and pursuer be $u,v$, respectively, and define $\lambda=\frac{u}{v} \ge 1$. Let the target move at constant speed along the $y$ axis on the line $(0,u\,t)$, with the pursuer following the radiodrome $\big(x(t),y(t)\big)$ starting from $x_0=x(0), y_0=y(0)$.
(a)  The condition that the pursuer aims directly towards the target translates to:
$$
y' = \frac{y - u\,t}{x}
$$
The arc length covered by the pursuer with constant speed $v$ after time $t$ is:
$$
v\,t = \int_{x}^{x_0} \sqrt{1 + (y')^2} \;dx
$$
Eliminating $t$ between the previous equations:
$$
\frac{1}{\lambda}(y - x\,y') = \int_{x}^{x_0} \sqrt{1 + (y')^2} \;dx
$$
Taking the derivates in $x$:
$$
\require{cancel}
\frac{1}{\lambda}(\bcancel{y'} - \bcancel{y'} - x\,y'') = -\,\sqrt{1 + (y')^2}
$$
With $w = y'$ and after separation of variables:
$$
\frac{dw}{\sqrt{1+w^2}} = \frac{\lambda \,dx}{x}
$$
Integrating:
$$
\operatorname{arcsinh}\big(w(x)\big) = \lambda \ln(x) + C
$$
The initial conditions $x=x_0, w(x_0)=y'(x_0) = \frac{y0}{x0}$ give the integration constant:
$$
C = \operatorname{arcsinh}(y_0\,/\,x_0) - \lambda \ln(x_0)
$$
At this point $y(x)$ can be explicitly calculated as $\int w \,dx$, but that's not needed at the next step.
(b)  The condition $y'(x_1)=w(x_1)=0$ translates to:
$$
0 = \lambda \ln(x_1) + \operatorname{arcsinh}(y_0\,/\,x_0) - \lambda \ln(x_0)
$$
$$
\lambda \ln(x_1 / x_0) =  - \operatorname{arcsinh}(y_0\,/\,x_0)
$$
$$
(x_1 / x_0)^\lambda = e^{-\operatorname{arcsinh}(y_0\,/\,x_0)}
$$
$$
x_1 = \frac{x_0} {\left(\frac{y_0}{x_0} + \sqrt{1+\left(\frac{y_0}{x_0}\right)^2}\right)^{\frac{1}{\lambda}}} = 
\left(\frac{x_0^{\lambda+1}}{y_0 + \sqrt{x_0^2+y_0^2}}\right)^{\frac{1}{\lambda}}
$$
(c)  Particularizing the general equations for $n$ lions and the limit case $x_1=1$:
$$
\begin{cases}
x_0 = 200 \,\sin\left(\frac{\pi}{n}\right) \\
y_0 = 200 \,\cos\left(\frac{\pi}{n}\right) \\
\lambda = 10
\end{cases}
$$
gives:
$$
200 \;\sqrt[10]{\frac{\sin^{11}(\pi/n)}{1 + \cos(\pi/n)}} = 1
$$
which solves numerically to:
$$
n \simeq 364.439
$$
It follows that $364$ lions allow an escape path, while $365$ will corner the target.
 [ EDIT ]   As a side note, using the first order approximations $\sin \alpha \sim \alpha$ and $\cos \alpha \sim 1$ for small $\alpha$ gives an approximation for the solution good to $2$ decimal places:

$$
n \;\approx\; \pi \; \sqrt[11]{\frac{200^{10}}{2}} \;\simeq\; 364.442
$$
A: First, some considerations. Both your body size and the lions' body size don't matter since all of them are considered points. Also the 1-meter limit around each lion acts like a circular force field around it, so if you enter its reach you're literally dead meat.
The lions are equally spaced around a circle of radius 200 meters, then you may consider they're exactly located at the vertices of a regular polygon with $X$ sides; also the lions are dumb since they do not cooperate and their trajectory from the starting position is a straight line toward the center, as show in figure 1 below.

Figure 1: your starting point $Z$ and two adjacent lions $L_{1}$ and $L_{2}$.
Since the lions are equally spaced, your best strategy is to take the shortest path out of the circular field, i.e., a radius. Take the line coincident to the bisector angle equally distant between two adjacent lions' trajectory lines.
Notice the lions trajectories do not cover all the field area, they leave some gap behind; the dangerous moment only occurs when their "force field" trajectory touch each other at the exact moment they also reach your trajectory. It means there is an encounter point $E$ intersecting your trajectory and their (represented as dashed lines) as shown in figure 2 below.

Figure 2: lions' trajectories and the encounter point $E$.
As you can imagine, the encounter point $E$ actually is the win condition for your freedom. Beyond that point you're safe because the lions cannot reach your trajectory anymore. And because the lions' speed is relatively one tenth of yours, the encounter point $E$ is always the same (with fixed values of $X$ and $\theta$). So keep in mind that you can run a distance ten times greater than a lion at the same period of time.
The initial question about the possible maximum value of $X$ has an immediate trivial answer given the 200 meters radius and the 1-meter force field radius: it's $X\leq628$, equivalent to the worst case scenario where the "force fields" are in touch from the beginning, result of the perimeter divided equally among 2-meters between lions (1-force-field length plus another), leaving no space to escape.
\begin{equation*}
p=2\cdot\pi\cdot r
\end{equation*}
\begin{equation*}
p=2\cdot\pi\cdot200
\end{equation*}
\begin{equation*}
p=1256.64\;m
\end{equation*}
\begin{equation*}
X=\frac{p}{2}=\frac{1256.64}{2}=628.319
\end{equation*}
\begin{equation*}
X\leq628
\end{equation*}
But this answer can be improved, since any value less than $628$ would leave some gap between the lions, right? Well, not exactly. Since you must run a lot before the gap closes.
Now imagine the minimum scenario where you can escape. Then you reach the escape point $E$ almost at the same time the lion's force field does. The following figure 3 is almost self-explicative:

Figure 3: a minimum narrow escape.
Remember you run ten times faster than any lion. Thus your escape distance $x=\overline{ZE}$ takes the same time a fat lion takes to walk the distance $\dfrac{x}{10}=\overline{GL_{1}}$
$ZGE$ is a right triangle such that $\overline{ZG}=r-\dfrac{x}{10}$. Thus it's possible to find your minimal escape distance,
\begin{equation*}
\left(\overline{ZE}\right)^{2}=\left(\overline{GE}\right)^{2}+\left(\overline{ZG}\right)^{2}
\end{equation*}
\begin{equation*}
x^{2}=1^{2}+\left(r-\dfrac{x}{10}\right)^{2}
\end{equation*}
\begin{equation*}
x^{2}=1+200^{2}-40x+\dfrac{x^{2}}{100}
\end{equation*}
\begin{equation*}
x^{2}-\dfrac{x^{2}}{100}+40x-40001=0
\end{equation*}
has a positive root $x=181.82068\;m$ is your escape distance.
With this value, you can find the minimal distance between two adjacent lions such that you can escape.
\begin{equation*}
\sin\dfrac{\theta}{2}=\dfrac{\overline{GE}}{\overline{ZE}}=\dfrac{\overline{FL_{1}}}{\overline{ZL_{1}}}
\end{equation*}
\begin{equation*}
\dfrac{1}{x}=\dfrac{a/2}{r}
\end{equation*}
\begin{equation*}
a=\dfrac{2r}{x}
\end{equation*}
\begin{equation*}
a=\dfrac{2\cdot200}{181.82068}
\end{equation*}
\begin{equation*}
a=2.19997\;m
\end{equation*}
this is the distance between lions in the minimum scenario where you can escape (approximately 2-force-fields plus $20\;cm$, remember you're point size).
Now look at the angle
\begin{equation*}
\sin\dfrac{\theta}{2}=\dfrac{1}{x}
\end{equation*}
\begin{equation*}
\dfrac{\theta}{2}=\sin^{-1}\left(\dfrac{1}{x}\right)
\end{equation*}
\begin{equation*}
\theta=2\cdot\sin^{-1}\left(\dfrac{1}{181.82068}\right)
\end{equation*}
\begin{equation*}
\theta=0.630248^{\circ}
\end{equation*}
Thus $X$ is the result of the division
\begin{equation*}
X=\dfrac{360^{\circ}}{\theta}=\dfrac{360^{\circ}}{0.630248^{\circ}}=571.204
\end{equation*}
Therefore, $X\leq571$ is the improved value.
